Introduction to Probability and Statistics for Engineers and Scientists

328 Chapter 8:Hypothesis Testing

whereXis a hypergeometric random variable with probability mass function

P{X=i}=

(

n 1

i

)(

n 2

k−i

)

(

n 1 +n 2

k

) i=0, 1,...,k (8.6.2)

In other words, this test will call for rejection if the significance level is at least as large as
thep-value given by

p-value=2 min(P{X≤x 1 },P{X≥x 1 }) (8.6.3)

This is called theFisher-Irwin test.

COMPUTATIONS FOR THE FISHER-IRWIN TEST

To utilize the Fisher-Irwin test, we need to be able to compute the hypergeometric
distribution function. To do so, note that withXhaving mass function Equation 8.6.2,

P{X=i+ 1 }

P{X=i}

=

(

n 1

i+ 1

)(

n 2

k−i− 1

)

(

n 1

i

)(

n 2

k−i

) (8.6.4)

=

(n 1 −i)(k−i)

(i+1)(n 2 −k+i+1)

(8.6.5)

where the verification of the final equality is left as an exercise.
Program 8.6.1 uses the preceding identity to compute thep-value of the data for the
Fisher-Irwin test of the equality of two Bernoulli probabilities. The program will work
best if the Bernoulli outcome that is called unsuccessful (or defective) is the one whose
probability is less than .5. For instance, if over half the items produced are defective, then
rather than testing that the defect probability is the same in both samples, one should test
that the probability of producing an acceptable item is the same in both samples.

EXAMPLE 8.6d Suppose that method 1 resulted in 20 unacceptable transistors out of 100
produced; whereas method 2 resulted in 12 unacceptable transistors out of 100 produced.
Can we conclude from this, at the 10 percent level of significance, that the two methods
are equivalent?

SOLUTION Upon running Program 8.6.1, we obtain that

p-value=.1763

Hence, the hypothesis that the two methods are equivalent cannot be rejected. ■